1 - 4 El APPLICATIONS OF DIFFERENTIATION 4.1 Maximum and...

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Unformatted text preview: 4 El APPLICATIONS OF DIFFERENTIATION 4.1 Maximum and Minimum Values \ 1. A function f has an absolute minimum at cc : c if f(c) is the smallest function value on the entire domain of f. whereas f has a local minimum at c if f(c) is the smallest function value when w is near c. 2. (a) The Extreme Value Theorem (b) See the Closed Interval Method. 3. Absolute maximum at b; absolute minimum at d; local maxima at b and 6; local minima at d and s; neither a maximum nor a minimum at a. c. r. and t. 4. Absolute maximum at e; absolute minimum at t; local maxima at c. e. and 8: local minima at b. c. d, and r; neither a maximum nor a minimum at a. 5. Absolute maximum value is f(4) : 4; absolute minimum value is f(7) 2 0; local maximum values are f(4) = 4 and f(6) = 3; local minimum values are f(2) = 1 and f(5) 2 2. 6. Absolute maximum value is f(8) : 5; absolute minimum value is f(2) : 0; local maximum values are f(1) = 2. f(4) : 4, and f(6) = 3; local minimum values are f(2) : 0. f(5) = 2. and f(7) = 1. 7. Absolute minimum at 2. absolute maximum at 3. 8. Absolute minimum at l. absolute maximum at 5. local minimum at 4 local maximum at 2. local minimum at 4 9. Absolute maximum at 5. absolute minimum at 2. 10. f has no local maximum or minimum, but 2 and local maximum at 3. local minima at 2 and 4 4 are critical numbers 263 264 CI CHAPTEIM APPLICATIONS OF DIFFERENTIATION 11. (a) y (b) 12. (a) Note that a local maximum (b) cannot occur at an endpoint. Note: By the Extreme Value Theorem. f must not be continuous. 13. (a) Note: By the Extreme Value Theorem, f must not (b) be continuous; because if it were. it would attain an absolute minimum. (b) 14. (a) 15. f(:c) : 8 — 331m 2 1. Absolute maximum f(1) : 5; no local maximum. No absolute or local minimum. 17. flat) 2 m2, 0 < m < 2. N0 absolute or local maximum or minimum value. 19. f(:c) : $2. 0 g m < 2. Absolute minimum f(0) = 0; no local minimum. No absolute or local maximum. 21. f(z) = $2. ~3 S x S 2. Absolute maximum f(—3) : 9. No local maximum Absolute and local minimum f(0) 2 0. (’19) y SECTION 4.1 MAXIMUM AND MINIMUM VALUES C 265 16. f(x) = 3 — 2m. x g 5. Absolute minimum f(5) = *7; no local minimum. No absolute or local maximum. 18. f(ac) : m2. 0 < at S 2. Absolute maximum f(2) : 4: no local maximum. No absolute or local minimum. 20. f(x) 2 1:2. 0 g .7: S 2. Absolute maximum f(2) = 4. Absolute minimum f(0) = 0. No local maximum or minimum. y 2. f(x) : 1+ (av +1)2. —2 S a: < 5. No absolute or local maximum. Absolute and local minimum f(—1):1. 265 3 CHAPTEFM APPLICATIONS OF DIFFERENTIATION 23. f(t) : 1/t. 0 < t < 1. No maximum or minimum. 25. f(0) : sin 0. —27r S 0 S 27r. Absolute and local maxima f(-37") = = 1. Absolute and local minimaf fig) : = —1. y 277 ’27 27- f(:c) = 1 — Absolute maximum f(0) : 1; no local maximum. No absolute or local minimum. 29 1—36 if0£zc<2 .flfli‘2m—4 fi2§mg3 Absolute maximum f (3) = 2; no local maximum. No absolute or local minimum. __._______,____.____________________________________________ 24. f(t) : 1/t.0 < t S 1. Absolute minimum f(1) : 1; no local minimum. No local or absolute maximum. 26. f(6) = tan 6, —§ 3 6 < Absolute minimum f(—§) : —1; no local minimum. No absolute or local maximum. (7. 28. f(:c) : er. No absolute or local maximum or minimum value. fl if—1§x<0 2—fi finggl safim:{ Absolute and local maximum f (0) : 2. N0 absolute or local minimum. SECTION 4.1 MAXIMUM AND MINIMUM VALUES 267 31. f(x) 2 5:52 +41: 2 f’(ac) 2 101: +4. f’(:c) 2 0 2> x 2 —§. so ~§ is the only critical number. 32.f(a:)2933+a:2—:c 2> f’(x)23x2+2;c—1. f’(w)20 2> (:c+1)(3:c—1)20 2> $2—1. These are the only critical numbers. A 3 . 33. flat) 2 $3 + 3502 — 24m => f’(m) = 3m2 + 63: 2 24 : 3(122 + 2x 2 8). f'(m) 2 0 2 3(w + 4)(93 — 2) 2 O 2> w 2 —4. 2. These are the only critical numbers. 34.f(;r)2:1:3+a:2+ac 2> f’(m)23m2+2$+1. f/(sc)20 2> 3$2+2z+120 2 a: 2 Neither of these is a real number. Thus. there are no critical numbers. 35. 5(t) = 32:4 + 4:3 — 6252 i s'(t)212t3 +12:2 212mm: 0 :5 12t(t2 + t — 1) 2 : 0 or t2 + t — 1 2 0. Using the quadratic formula to solve the latter equation gives us a / 2 _ _ 7 2 :: t : 1i 1 4(1)( 1) : 1i «5 m 0.618. —1.618. The three critical numbers are 0. 2(1) 2 2 z+1 I (22+z+1)1~(z+1)(2z+1) 222—22 35_ : _— :> : — 2 0 <=> flz) 22+z+1 HZ) (22+z+1)2 (z2+z+1)2 z(z + 2) 2 0 2> z 2 0. —2 are the critical numbers. (Note that 22 + z + 1 2 0 since the discriminant < 0.) 2x+3 if2m+320 2 ifzc>—g 37. g(x) 2 |2$ + 3| 2 _ 2> g’($) 2 —(2;c+3) lf2x+3<0 22 ifx<—§ g'(:c) is never 0. but g'(;c) does not exist form 2 —%. so 2% is the only critical number. _ , _ _ _ :6 +2 38. g($) 2m1/3 —;z: 2/3 :5 g (x) : gx 2/3+ gm 5/3 = gm 5/3(x+2) : 3365/3. g'(e2) 2 0 and g’(0) does not exist. but 0 is not in the domain ofg. so the only critical number is 22. 39. g(t) 2 52‘.”3 + 235/3 2> g’(t) 2 339271” + 319/3. g’(0) does not exist. sot 2 0 is a critical number. g’(t) 2 §t_1/3(2 + t): 0 <=> t 2 —2. so t 2 —2 is also acritical number. 1 40. g(t) 2 \/£(1—t) 2t1/2 — t3/2 2 g’(t) 2 2 fl — g’(0) does not exist. sot 2 0 is acritical number. ‘ 1— 3t 2%? 41. F(x) 2 x4/5(x — 4)2 2> F’(m) 2x4/5‘2(a:*4)+(x—4)2~gx’1/52%m_1/5(z~4)[5-$t2+($—4)~4] mv414x—16 216—4 730—8 = % 2 % 2 OWhenx 2 4. §;and F'(0) does not exist. 0 2 g'(t) 2> t 2 . sot 2 is also a critical number. wIH rah—i Critical numbers are 0. g. 4. 42. C(30) 2 {97:02 — m 2 G’(a:) 2 an? ~ at) 72/3 (2m ~ 1). G’(w) does not exist when $2 — m 2 0. that is. when :1: 2 Oor 1. G'(x) 2 0 <2 2m — 1 2 0 <2 as 2 So the critical numbers arem 2 0.1.1. N ___/ 268 43. 44. 45. 46. 47. 48. 49. 50. 51. 52. 53. 3 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION f(6):2cos6+sin26 => f’(6) —2sin6+2sin6cos6. If’(6):0 :> 2sin0(cosa_1):0 :> sin6 = 0 or cos6 : 1 :> 6 : mt (n an integer) or 6 : 2n7r. The solutions 6 : mr include the solutions 6 : 2n7r. so the critical numbers are 6 = 71,77. 9(6) : 46 — tan6 => g'(6) : 4 — sec26. g'(6) = 0 :> sec26 : 4 :> sec6 : :I:2 => c0s6 = :I:% => 6 = g + 2n7r. 5?" + 27m. + 21177. and 4?" + 2mr are critical numbers. Note: The values of 6 that make 9' (6) undefined are not in the domain of g. f(cc) : xlncc :> = + (11156) - 1 = 11135 +1. f'(x) = 0 <:> Inst 2 ~1 <=> z = 6-1 : 1/e. Therefore. the only critical number is as = 1/6. f(m) : $629” => f’(:c) : M26”) + 629” : e2z(2:c + 1). Since e21 is never 0. we have f’(m) : 0 only when 2m + 1 = O <=> ac : —%. So A% is the only critical number. f(m) : 3w2 # 12m + 5. [0.3]. f'(:c) : 6x — 12 : 0 <=> ac : 2. Applying the Closed Interval Method. we find that f(0) = 5. f(2) #7. and f(3) absolute minimum value. = —4. So f(O) : 5 is the absolute maximum value and f(2) : —7 is the = 3:3 ~ 3x+1. [0.3]. f’(:c) = 39:2 ~ 3 : 0 <=> a: = :I:1.but *1 is not in [0.3]. f(O) : 1. f(1) = *1. and f(3) : 19. So f(3) = 19 is the absolute maximum value and f(1) : —1 is the absolute minimum value. f(m) : 2m3 — 3:02 — 12$ + 1. [72,3]. f’(a:) = 6m2 v 69: — 12 = 66:2 — an — 2) : 6(30 — 2)($ + 1) = 0 <=> cc 2 2. —1. f(—2) : #3.f(—1): 8. f(2) : —19. and f(3) = 78. So f(#1) = 8 is the absolute maximum value and f(2) : —19 is the absolute minimum value. f(cc) : x3 — 6:02 + 9x + 2. [—1.4]. f'(x) : 32:2 —12:c + 9 : 3(362 — 4x + 3) : 3(1‘ 41mg — 3) : 0 ¢> :c : 1,3. f(—1) —14. f(1) = 6. f(3) = 2. and f(4) = 6. So f(1) 2 f(4) = 6 is the absolute maximum value and f(—1) = —14 is the absolute minimum value. f(x):w4—2x2+3.[i2.3]. f’(m):4x3~4:c=4w(sc2—1):4$($+1)($#1):0 <2; :E:71.0,1. f(-2)-11~f( 1) 2~f(0) 3.f(1) 21(3) f(:l:1) = 2 is the absolute minimum value. 66. So f(3) : 66 is the absolute maximum value and 2 flat) = (962 — 1)3~ [*LQI‘ NSC) : 3(932 41) (250) f(:I:1) : 0. f(O) : —1. and f(2) : 27. So f(2) : 27 is the absolute maximum value and f(O) : *1 is the :603(:c+1)2(:c~1)2 :0 <=> 30:71.0,1. absolute minimum value. (:52 +1) # m(2x) 2 f/(ic) ' (x2 +1)2 .f<2) : % So f(1) =% # 1—ac _($2+1)2 is the absolute maximum value and f (0) z 0 is the absolute ac . —() 42> m=i1.but—1isnotin[0,2]. +1 ax) : w. [0.2]. no) 2 0. f(1) = % minimum value. 2 2 2 ac — 4 (m + 4X29?) — (w — 4)(2:c) 16a: 0 _ g — ’ a - ————— 2 <:> ac # O. 54' _ m2 + 4~ I 4‘4]. f (m2 + 412 (902 + 4)2 f(i4) : é—g : and f(O) : —1. So f(i4) : g is the absolute maximum value and f(O) : —1 is the absolute minimum value. SECTION 4.1 MAXIMUM AND MINIMUM VALUES E 269 55. f(t) = tx/4~t2. [—1.2]. _ _t2 —t2 + (4 — t2) 4 — 2t2 ’t:t-l4it2 1/2~2t 4421/21: +I/4—t2:‘: . 57. 59. 60. 61. 62. .f(t) : %(8 i t). [078} f(t) : 8751/3 — W3 :> f’(t) : gr?” — 3151/3 : §t*2/3(2 i t) : f’(t) — 0 4 2t2 g 0 —> t2 i 2 —> t— ifi. butt 2 ix/i is not in the given interval, [—1,2]. f'(t) does not exist if4 — t2 : 0 => t: i2. but 72 is not in the given interval. f(—1) = i\/§. 2 2, and f(2) : 0. So = 2 is the absolute maximum value and f(—1) fix/g is the absolute minimum value. f’(t) : 0 :> t: 2. f’(t) does not exist ift : 0. f(0) = 0. f(2) = 6 \3/5 a 7.56. and f(8) : 0. So f(2) : 6 \3/5 is the absolute maximum value and f(0) : f(8) : O is the absolute minimum value. . . s'nx f’(m)=cosm—s1n$:0 <:> smxzcosx => 1 f(:c) : sinw + cos ac. [0. => W|=l COS {I} i f(0):1~f(§) x/iz 1.41.f(§) : £211 absolute maximum value and f(0) : 1 is the absolute minimum value. tanx:1 => 95: z1.37.30f(g):\/§Isthe .f($)::ci2cosm. [—7r,7r]. f’(a:):1+2sinm=0 <:> sinm=—% <=> ac: 7%" —§. f(~7r) : 2 — 7r z —1.14.f(—5%) = fa 5?" a: 4.886. flag) : 7% i fix —2.26. f(7r) : 71' + 2 m 5.14. So f(7r) 2 71' + 2 is the absolute maximum value and flig) : 73 ~ \/§ is the absolute 6 minimum value. : ace—z. [07 2]. f'(m) : Cid—€02) + e”: : 671(1— 26) : 0 <=> 9: =1. f(0) : 0. f(1) : 6—1 : l/e % 0.37, f(2) : 2/e2 a: 0.27. So f(1) : 1/6 is the absolute maximum value and f(0) : 0 is the absolute minimum value. lnx , x(1/w)—lnm 1~lnx f(w)=T= m2 2 <=>1—ln$:0 <=> lnac=1 <=> x:e. f(1) : 0/1 : 0. f(e) : l/e z 0.368. f(3) : (1n 3)/3 m 0.366. So f(e) and f(1) : 0 is the absolute minimum value. : l/e is the absolute maximum value x—3 m in the domain off. f(1) : 1. f(3) : 3 — 31113 x —0.296. f(4) : 4 — 3ln4 % 70.159.Sof(1):1isthe = x — 31mm [1.4]. f'(;v) : 1* : O <=> m : 3. f’ does not exist forsc : 0. but 0 is not RICO absolute maximum value and f(3) : 3 — 3ln3 x —0.296 is the absolute minimum value. 2 1_2—em 61 ms) : 6‘2”. [01]. f’(x) : e‘I(—]) ~ e_2z(~2) = 622 621 :1: : m2 m 0.69. f(0) : 0. f(ln 2) : e— ‘1‘2 — e41” : (emf1 * (51“2)‘2 2‘1 — 2’2 : l 2 f(1) : 6—1 — e’2 x 0.233. So f(ln 2) : i is the absolute maximum value and f(0) : 0 is the absolute minimum value. / 270 C CHAPTERlI APPLICATIONS OF DIFFERENTIATION 63.f(z):x“(1—sc)b.03x31,a>0.b>0. f’(x) : w“ - b(1 i :c)b_1(—1)+(1 — 1:)b - (wail : w“_1(17 ac)b_1[:c-b(—1)+(1 — x) - a] : zed-1U W cc)b'1(a * an: i bat) At the endpoints. we have f(0) : f(1) = 0 [the minimum value of f]. In the interval (0,1). f’(:17) : 0 ¢> m _ a _ a + I). f( a )_( a )a(1 a >b_ a“ a+bia b_ a“ bb _ aabb a+b a+b a+b (a—I—b)“ a—I—b F(a+b)“ (a+b)bi(a+b)a+b' a aabb So f > = ———afi is the absolute maximum value. a + b (a + b) 64. 10 We see that f’(:c) = 0 at about a: = 0.0 and 2.0. and that f’(w) does not exist at about x = #07. 1.0. and 27. so the critical numbers of f f are about 70.7. 0.0. 1.0. 2.0. and 2.7 65. (a) From the graph. it appears that the absolute maximum value is about f(—1.63) = 9.71. and the absolute minimum value is about f(1.63) = 77.71. These values make sense because the graph is symmetric about the point (0. 1). (y = :63 — 82c is symmetric about the origin.) (b)f(x)::c3—8w+1 :> f’(a:):3m2—8.Sof’(a:):0 :> m:i\/§ 3 Mam/g) (: g) —8(i\/§)+1:i§\/§¢8\/§+1 :i%\/§+1:1—3—2§@ [minimum] or %\/§+1=1+§249@ [maximum] (From the graph. we see that the extreme values do not occur at the endpoints.) II 2 From the graph. it appears that the absolute maximum value is about fl f(—0.58) : 1.47. and the absolute minimum value is about - f(#1) : f(0) : 1.00; that is. at both endpoints. (b) fa) : ems-I ;» f'(;£) : era-1cm? — 1). So f’(x) : Oon [—1.0] :> :3 : — 1/3. f(—1): f(0) : 1 (minima) and f(—t/1/3) : (“g/9+3” : e2‘/§/9 (maximum). 56. (a) SECTION 4.1 MAXIMUM AND MINIMUM VALUES 3 271 67. (a) 04 From the graph. it appears that the absolute maximum value is about f(0.75) 2 0.32. and the absolute minimum value is f(0) 2 f(1) 2 0; that is. at both endpoints. I 1_2 (2—2362) + (Zr—23:2) Elm—42:2 (“Mlflw‘fl : f($):$'2\/T—:2+”‘m2’ 2m ‘2m' Sof’(x)20 :> 3x24m220 => $(3—4x)20 :> 9:20or431.f(0)2f(1)20[minimum], andflg) : § %— ($2 : ¥ [maximum]. 68. (a) 08 From the graph. it appears that the absolute maximum value is about ' f(5.76) 2 0.58. and the absolute minimum value is about 0 L ‘ 2,, f(3.67) 2 ~0.58. —0i8 (1))“ )fi cosa: :> fa) _ (2+sinm)(—sinx)—(cosx)(cosm) —1—2sina: x _ 2 + sina: (2 + sin :c)2 (2 + sin :5)? So f’(ac) — 0 2,» sinx — —% —> a: — 7F” or 11?". Now “76%) 2 _;//_322 2 ~% [minimum]. and f(“T") 2 gg 2 % [maximum]. 69. The density is defined as p 2 Elms: 2 (in g/cm3). But a critical point ofp will also be a critical point v um d of V [since 3—; 2 —1000V’2% and V is never 0]. and V is easier to differentiate than p. V(T) : 999.87 2 0.06426T + 0.0085043T2 — 0‘0000679T3 => V/(T) 2 —0.06426 + 0.0170086T — 0.0002037T2i Setting this equal to O and using the quadratic formula to ‘ 2 2 . . find T, we get T ‘ 0.0170086 2t x/0.0I70086 4 00002037 0.06426 2(—0.0002037) z 3.96650C or 79.53180C. Since we are only interested in the region 0°C 3 T S 300C. we check the density p at the endpoints and at 3.966500 1000 1000 1000 0 % @100013; 30 z ‘ m _ ; ‘ z a I . ( ) 999.87 p( ) 10037628 099625 p(3 9665) 9997447 1000255 So water has its maximum density at about 3.9665”Ci / 272 I: CHAPTER 4 APPLICATIONS OF DIFFERENTIATION 70 F : MW E i (,u sinO + cos 9)(0) — MWW c080 - sin 0) _ —[LW(M c050 — sin 0) usinO +cos€ d0 (,usinB—I—cosé)2 (psin9+c056)2 I dF ‘ So E = 0 2) pcosfi sin0 * 0 p — ’ tan 0. Substituting tanB for ,u in F gives us F: (tanO)W _ Wtan6 _ Wtan6c056 f Wsin6 #W . 6 (tan 9) sin 0 + cos 0 T sin2 6' T sin2 6 + cos2 6 T 1 _ 8m ' + (3036 cos 0 If tame : p. then sin 9 : L (see the figure). so F : ——M——W. We #2 + 1 i/u2 + 1 compare this with the value of F at the endpoints: F(0) : MW and = W. M u u Now because 3 1 and —— g ,u, we have that ————W u2+1 i/u2+1 «02H 71. 72. (a) is less than or equal to each of F(0) and Hence. W is the absolute minimum value of and it u2+1 occurs when tanQ = 11.. We apply the Closed Interval Method to the continuous function I(t) = 0000090452‘,5 + 0.001435%4 — 0.06561t3 + 0.45982:2 ‘ 0.6270t + 99.33 on [010]. Its derivative is I’(t) : 0.00045225t4 + 0.005752t3 ~ 0.19683t2 + 0.919615 — 0.6270. Since 1' exists for all t. the only critical numbers of I occur when I ' (t) : 0. We use a root—finder on a computer algebra system (or a graphing device) to find that I'(t) : 0 when t % —29.7186. 0.8231. 5.1309. or 11.0459. but only the second and third roots lie in the interval [0. 10]. The values of I at these critical numbers are I(0.8231) % 99.09 and I(5.1309) % 100.67. The values of I at the endpoints of the interval are 1(0) : 99.33 and I (10) m 96.86. Comparing these four numbers. we see that food was most expensive at t a“ 5.1309 (corresponding roughly to August. 1989) and cheapest at t : 10 (midyear I994). 4200 The equation of the graph in the figure is v(t) : 000146253 — 0.11553132 + 24.98169t ‘ 21.26872. 0 125 (b) a(t) : 0’05) 2 0.004329:2 0.23100: + 24.98169 :> a'(t) : 0.0087615 — 0.23106. do) = 0 : t1 : fig: m 26.4. (1(0) m 24.98. a(t1) R: 21.93. and 0(125) :3 64.54. The maximum acceleration is about 64.5 ft/s2 and the minimum acceleration is about 21.93 ft/sg. SECTION 4.1 MAXIMUM AND MINIMUM VALUES 3 273 73. (a) v(r) 2 k(ro ~ r)r2 2 kror2 — MS 2> 1/(7‘) 2 2km?" — 3kr2. 1/0") 2 0 2> kr(2ro — 3r) 2 0 2> 7" 2 O or gm (but 0 is not in the interval). Evaluating v at gm. gm, and m. we get vGro) 2 ékrg. v(§r0) 2 2%]97‘8. and v(r0) 2 0. Since 237 > 31;. 17 attains its maximum value at r 2 gm. This supports the statement in the text. 4 3 7197]). (b) From part (a), the maximum value ofv is 2— (C) 74. 9(a) 2 2 + (ac 2 5)3 => g'(93) 2 3(a: ‘ 5)2 2> g'(5) 2 0. so 5 is a critical number. But 9(5) 2 2 and 9 takes on values > 2 and values < 2 in any open interval containing 5. so 9 does not have a local maximum or minimum at 5. 75. f(x) 2 $101+ 3351 + at +1 => f’(z) 2 10150100 + 51x50 +1 2 1 for all 5v. so f’($) 2 0 has no solution. Thus. f(.23) has no critical number. so f(a:) can have no local maximum or minimum. 76. Suppose that f has a minimum value at c. so f($) 2 f(c) for all :6 near c. Then 9(50) 2 if(:c) < —f(c) 2 9(0) for all at near c. so 9(90) has a maximum value at c. 77. If f has a local minimum at c. then g(m) 2 —f($) has a local maximum at c. so g’(c) 2 0 by the case of Fermat's Theorem proved in the text. Thus. f'(c) 2 —g'(c) 2 0. 78. (a) fix) 2 (1503 + M? + can + d. a 76 0. So f’(m) 2 3ax2 + 2179: + c is a quadratic and hence has either 2. 1. or 0 real roots. so f(x) has either 2. 1 or 0 critical numbers. Case (i) (2 critical numbers): Case ( ii ) (1 critical number): Case (iii) (no critical number): f<x> : x3 — 3w 2 fix) : 6” => are) : H + 3x ;» f'(ac) 2312—38037: —1.1 f’(w) 23m2.sow20 f'(a:) 23$2+3. are critical numbers. is the only critical number. so there are no real roots. y (b) Since there are at most two critical numbers. it can have at most two local extreme values and by (i) this can occur. By (iii) it can have no local extreme value. However. ifthere is only one critical number. then there is no local extreme value. ...
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This note was uploaded on 12/08/2009 for the course MATH 101 taught by Professor Dr.tahir during the Fall '08 term at King Fahd University of Petroleum & Minerals.

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1 - 4 El APPLICATIONS OF DIFFERENTIATION 4.1 Maximum and...

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